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\title{\heiti\zihao{2} 习题4.7}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{计算以下极限:}
\subsection{$\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}=\lim _{x \rightarrow 0} \frac{1-\cos x}{3x^{2}}=\lim _{x \rightarrow 0} \frac{\frac{x^{2}}{2}}{3x^{2}}=\frac{1}{6}$$

\subsection{$\lim _{x \rightarrow 0} \frac{\tan x-x}{x^{3}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0} \frac{\tan x-x}{x^{3}}=\lim _{x \rightarrow 0} \frac{\sec^{2}x-1}{3x^{2}}=\lim _{x \rightarrow 0} \frac{\tan^{2}x}{3x^{2}}=\frac{1}{3}$$

\subsection{$\lim _{x \rightarrow 0} \frac{x-\frac{x^{2}}{2}-\ln (1+x)}{x^{3}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0} \frac{x-\frac{x^{2}}{2}-\ln (1+x)}{x^{3}}=\lim _{x \rightarrow 0} \frac{1-x-\frac{1}{1+x} }{3x^{2}}=\lim _{x \rightarrow 0} \frac{1-x-\frac{1}{1+x} }{3x^{2}}=-\frac{1}{3}$$

\subsection{$\lim _{x \rightarrow 0+} \frac{\ln (\tan 5 x)}{\ln (\tan 2 x)}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0+} \frac{\ln (\tan 5 x)}{\ln (\tan 2 x)}=\lim _{x \rightarrow 0+} \frac{\tan 2 x}{\tan 5 x}\frac{\cos^2 2 x}{\cos^2 5 x}\frac{5}{2}=1$$

\subsection{$\lim _{x \rightarrow 1} \frac{\ln (\cos (x-1))}{1-\sin \frac{\pi x}{2}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 1} \frac{\ln (\cos (x-1))}{1-\sin \frac{\pi x}{2}}=\lim _{x \rightarrow 1} \frac{-\tan(x-1)}{-\cos \frac{\pi x}{2}\frac{\pi}{2}}=\lim _{x \rightarrow 1} -\frac{4\sec^2(x-1)}{\sin \frac{\pi x}{2}\pi^{2}}=-\frac{4}{\pi^2}$$

\subsection{$\lim _{x \rightarrow 1} \frac{x-x^{x}}{1-x+\ln x}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 1} \frac{x-x^{x}}{1-x+\ln x}=\lim _{x \rightarrow 1} \frac{1-x^{x}(\ln x+1)}{-1+\frac{1}{x} }=\lim _{x \rightarrow 1} \frac{x^{x}(\ln x+1)^2+x^{x-1}}{\frac{1}{x^2} }=1+1=2$$

\subsection{$\lim _{x \rightarrow 0} \frac{e^{x}-\sqrt{1+2 x}}{\ln \left(1+x^{2}\right)}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0} \frac{e^{x}-\sqrt{1+2 x}}{\ln \left(1+x^{2}\right)}=\lim _{x \rightarrow 0} \frac{e^{x}-\frac{1}{\sqrt{1+2 x}}}{\frac{2x}{1+x^{2}}}=\lim _{x \rightarrow 0} \frac{e^{x}+\frac{1}{{1+2 x}^{\frac{3}{2}}}}{\frac{2(1-x^2)}{(1+x^{2})^2}}=\frac{2}{2}=1$$

\subsection{$\lim _{x \rightarrow 0} \sqrt{1-x^{2}} \cot \left(\frac{x}{2} \sqrt{\frac{1-x}{1+x}}\right)$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0} \sqrt{1-x^{2}} \cot \left(\frac{x}{2} \sqrt{\frac{1-x}{1+x}}\right)=\lim _{x \rightarrow 0}  \cot \left(\frac{x}{2} \right)=\infty$$

\subsection{$\lim _{x \rightarrow 0} x^{2} \mathrm{e}^{\frac{1}{x^{2}}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0} x^{2} \mathrm{e}^{\frac{1}{x^{2}}}=\lim _{x \rightarrow 0}  \frac{\mathrm{e}^{\frac{1}{x^{2}}}}{\frac {1}{x^{2}}}=\lim _{t \rightarrow +\infty}  \frac{\mathrm{e}^{t}}{t}=+\infty$$

\subsection{$\lim _{x \rightarrow \infty} x\left[\left(1+\frac{1}{x}\right)^{x}-\mathrm{e}\right]$}
\textbf{解}\quad
$$\lim _{x \rightarrow \infty} x\left[\left(1+\frac{1}{x}\right)^{x}-\mathrm{e}\right]=\lim _{x \rightarrow \infty} \frac{\left[\left(1+\frac{1}{x}\right)^{x}-\mathrm{e}\right]}{\frac{1}{x}}=\lim _{t \rightarrow 0} \frac{\left[\left(1+t\right)^{\frac{1}{t}}-\mathrm{e}\right]}{t}=\lim _{t \rightarrow 0} \mathrm{e}\frac{\left[t-\ln(1+t)\right]}{t^2(1+t)}=\frac{\mathrm{e}}{2}$$

\subsection{$\lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)=\lim _{x \rightarrow 0}\frac{x-\sin x}{x^2}=\lim _{x \rightarrow 0}\frac{1-\cos x}{2x}=0$$

\subsection{$\lim _{x \rightarrow-x}\left(\frac{1}{\sin ^{2} x}-\frac{1}{(x-\pi)^{2}}\right)$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim _{x \rightarrow\pi}\left(\frac{1}{\sin ^{2} x}-\frac{1}{(x-\pi)^{2}}\right) & =\lim _{t \rightarrow0}\left(\frac{1}{\sin ^{2} t}-\frac{1}{t^{2}}\right) \\
		                                                                                 & =\lim _{t \rightarrow0}\frac{t^2-\sin^2t}{t^{4}}                          \\
		                                                                                 & =\lim _{x \rightarrow 0} \frac{2 x-\sin 2 x}{4 x^{3}}                     \\
		                                                                                 & =\lim _{x \rightarrow 0} \frac{2-2 \cos 2 x}{12 x^{2}}                    \\
		                                                                                 & =\lim _{x \rightarrow 0} \frac{\sin 2 x}{6 x}=\frac{1}{3}
	\end{aligned}
$$

\subsection{$\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)=\lim _{x \rightarrow 0}\left(\frac{e^{x}-1-x}{x\left(e^{x}-1\right)}\right)=\lim _{x \rightarrow 0}\left(\frac{e^{x}-1}{e^{x}-1+e^{x} x}\right)=\lim _{x \rightarrow 0}\left(\frac{e^{x}}{e^{x} x+2 e^{x}}\right)=\frac{1}{2}$$

\subsection{$\lim _{x \rightarrow 0}(\cos \pi x)^{\frac{1}{x^{2}}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0}(\cos \pi x)^{\frac{1}{x^{2}}}=\lim _{x \rightarrow 0} e^{\frac{\ln \cos \pi x}{x^{2}}}=-\lim _{x \rightarrow 0} e^\frac{\pi \tan \pi x}{2 x}=e^{-\frac{\pi^{2}}{2}}$$

\subsection{$\lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^{2}}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^{2}}}=\lim _{x \rightarrow 0} e^{\frac{1}{x^{2}} \ln \left(\frac{\tan x}{x}\right)}$$

先求极限
$$
	\begin{aligned}
		\lim _{x \rightarrow 0} {\frac{1}{x^{2}} \ln \left(\frac{\tan x}{x}\right)} & =\lim _{x \rightarrow 0} \frac{\frac{x}{\tan x} \frac{x \sec ^{2} x-\tan x}{x^{2}}}{2 x} \\
		                                                                            & =\lim _{x \rightarrow 0} \frac{x \sec ^{2} x-\tan x}{2 x^{3}}                            \\
		                                                                            & =\lim _{x \rightarrow 0} \frac{2 x \sec ^{2} x \tan x+\sec ^{2} x-\sec ^{2} x}{6 x^{2}}  \\
		                                                                            & =\lim _{x \rightarrow 0} \frac{2 \sec ^{2} x}{6}                                         \\
		                                                                            & =\frac{1}{3}
	\end{aligned}
$$

故原极限为$e^{\frac{1}{3}}$

\subsection{$\lim _{x \rightarrow 0}\left[\frac{\ln (1+x)}{x}\right]^{\frac{1}{x}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0}\left[\frac{\ln (1+x)}{x}\right]^{\frac{1}{x}}=\lim _{x \rightarrow 0} e^{\frac{\ln \left(\frac{\ln (1+x)}{x}\right)}{x}}$$

先求极限

$$
	\begin{aligned}
		\lim _{x \rightarrow 0} {\frac{\ln \left(\frac{\ln (1+x)}{x}\right)}{x}} & =\lim _{x \rightarrow 0}\frac{\frac{\frac{x}{1+x}-\ln (1+x)}{x^{2}}}{\frac{\ln (1+x)}{x}} \\
		                                                                         & =\lim _{x \rightarrow 0}\frac{\frac{x}{1+x}-\ln (1+x)}{x^{2}}                             \\
		                                                                         & =\lim _{x \rightarrow 0}\frac{\frac{1}{(1+x)^{2}}-\frac{1}{1+x}}{2 x}                     \\
		                                                                         & =-\frac{1}{2}
	\end{aligned}
$$

所以原式极限为$e^{-\frac{1}{2}}$

\subsection{$\lim _{x \rightarrow 0}\left(\ln \frac{1}{x}\right)^{x}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0}\left(\ln \frac{1}{x}\right)^{x}=\lim _{x \rightarrow 0}e^{x\ln\left(\ln \frac{1}{x}\right)}$$

先求极限

$$
	\begin{aligned}
		\lim _{x \rightarrow 0} x \ln \left(\ln \frac{1}{x}\right) & =\lim _{x \rightarrow 0} \frac{\ln \left(\ln \frac{1}{x}\right)}{\frac{1}{x}} \\
		                                                           & =\lim _{t \rightarrow \infty} \frac{\ln (\ln t)}{t}                           \\
		                                                           & =\lim _{t \rightarrow \infty} \frac{\frac{1}{t}}{\ln t}                       \\
		                                                           & =0
	\end{aligned}
$$

所以原极限为$e^0=1$

\subsection{$\lim _{x \rightarrow 0}(\sin x)^{\tan x}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0}(\sin x)^{\tan x}=\lim _{x \rightarrow 0} e^{\tan x \ln \sin x}$$

先求极限

$$\lim _{x \rightarrow 0} \tan x \ln (\sin x)=\lim _{x \rightarrow 0} \frac{\ln \sin x}{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{\frac{\cos x}{\sin x}}{-\frac{1}{x^{2}}}=0$$

所以原极限为$e^0=1$

\subsection{$\lim _{x \rightarrow 0}(\tan x)^{\sin x}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0}(\tan x)^{\sin x}=\lim _{x \rightarrow 0} e^{\sin x \ln \tan x}$$

先求极限

$$\begin{array}{l}
		\lim _{x \rightarrow 0} \sin x \ln \tan x=\lim _{x \rightarrow 0} \frac{\ln \tan x}{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{\frac{\sin x}{\cos ^{3} x}}{-\frac{1}{x^{2}}}=0
	\end{array}$$

所以原极限为$e^0=1$

\subsection{$\lim _{x \rightarrow 0}\left[\frac{(1+x)^{\frac{1}{x}}}{\mathrm{e}}\right]^{\frac{1}{x}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0}\left[\frac{(1+x)^{\frac{1}{x}}}{\mathrm{e}}\right]^{\frac{1}{x}}=\lim _{x \rightarrow 0} e^{\frac{1}{x} \cdot \ln \frac{(1+x)^{\frac{1}{x}}}{e}}$$

先求极限

$$\lim _{x \rightarrow 0} \frac{\ln \left(\frac{(1+x)^{\frac{1}{x}}}{e}\right)}{x}=\lim _{x \rightarrow 0} \frac{\frac{1}{x} \ln (1+x)-1}{x}=\lim _{x \rightarrow 0} \frac{\ln (1+x)-x}{x^{2}}=\lim _{x \rightarrow 0} \frac{\frac{1}{x+1}-1}{2 x}=-\frac{1}{2}$$

所以原极限为$e^{-\frac{1}{2}}$

\subsection{$\lim _{x \rightarrow \infty}\left(\tan \frac{\pi x}{2 x+1}\right)^{\frac{1}{x}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow \infty}\left(\tan \frac{\pi x}{2 x+1}\right)^{\frac{1}{x}}=\lim _{x \rightarrow 0} \mathrm{e}^{\frac{\tan \frac{\pi x}{2 x+1}}{x}}=\mathrm{e}^{\lim \frac{\tan \frac{\pi x}{2 x+1}}{x}}$$

$$\begin{aligned}
		\lim _{x \rightarrow \infty} \frac{\tan \frac{\pi x}{2 x+1}}{x} & =\lim _{x \rightarrow \infty} \frac{\sec ^{2} \frac{\pi x}{2 x+1}}{\tan \frac{\pi x}{2 x+1}} \frac{\pi}{(2 x+1)^{2}}=\lim _{x \rightarrow \infty} \frac{\frac{\pi}{(2 x+1)^{2}}}{\cos \frac{\pi x}{2 x+1}} \\
		                                                                & =\lim _{x \rightarrow \infty} \frac{\frac{4 \pi}{(2 x+1)^{3}}}{\frac{\pi}{(2 x+1)^{2}} \sin \frac{\pi x}{2 x+1}}=\lim _{x \rightarrow \infty} \frac{\frac{4}{(2 x+1)}}{\sin \frac{\pi x}{2 x+1}}=0
	\end{aligned}$$

所以原式=$e^0=1 $

\section{设$a_{n}=\mathrm{e}-\left(1+\frac{1}{n}\right)^{n}, n=1,2, \cdots$,计算极限$\lim _{n \rightarrow \infty} n a_{n} .$}
\textbf{解}\quad
构造函数
$$
	\begin{aligned}
		f(x) & =\frac{e-(1+x)^{\frac{1}{x}}}{x}\lim _{x \rightarrow 0} f(x)                                              \\
		     & =\lim _{x \rightarrow 0}-e^{\frac{1}{x} \ln (1+x)}\left(-\frac{\ln (1+x)}{x^{2}}+\frac{1}{(1+x) x}\right) \\
		     & =\lim _{x \rightarrow 0}-e \frac{x-(1+x)\ln(1+x)}{x^{2}}                                                  \\
		     & =-e\lim _{x \rightarrow 0} \frac{-\ln (1+t)}{2 x}                                                         \\
		     & =\frac{e}{2}
	\end{aligned}
$$

所以由海涅定理知,数列极限为$\frac{e}{2}$

\section{设$a_{i}>0, i=1,2, \cdots, n $. 求极限 $\lim _{x \rightarrow a}\left(\frac{a_{1}^{x}+a_{2}^{x}+\cdots+a_{n}^{x}}{n}\right)^{\frac{1}{x}}$,  这里$a$取$0$或$\pm \infty $.}
\textbf{解}\quad
当$a=0$时,
$$
	\begin{aligned}
		\lim _{x \rightarrow 0}\left(\frac{a_{1}^{x}+a_{2}^{x}+\cdots+a_{n}^{x}}{n}\right)^{\frac{1}{x}} & =
		e^{\lim _{x \rightarrow 0}} \frac{\ln \frac{a_{1}^{x}+a_{2}^{x}+\cdots+a_{n}^{x}}{n}}{x}                                                                                                                                                                 \\
		                                                                                                 & =e^{\lim _{x \rightarrow 0}}\frac{\frac{a_{1}^{x} \ln a_{1}+a_{2}^{x} \ln a_{2}+\cdots+a_{n}^{x} \ln a_{n}}{a_{1}^{x}+a_{2}^{x}+\cdots+a_{n}^{x}}}{1} \\
		                                                                                                 & =e^{\lim _{x \rightarrow 0} \frac{\ln a_{1}+\ln a_{2}+\cdots+\ln a_{n}}{n}}                                                                           \\
		                                                                                                 & =\sqrt[n]{a_{1}a_{2}\cdots a_{n}}
	\end{aligned}
$$

当$a=+\infty$时,

$\lim _{x \rightarrow +\infty}\left(\frac{a_{1}^{x}+a_{2}^{x}+\cdots+a_{n}^{x}}{n}\right)^{\frac{1}{x}}=e^{\lim _{x \rightarrow \infty}\frac{1}{x}\ln \frac{a^x_1+a^x_2+\cdots+a^x_n}{n}}$

先求极限

$\lim _{x \rightarrow+\infty} \frac{1}{x}\left[\ln \left(a_{1}^{x}+a_{2}^{x}+\cdots+a_{n}^{x}\right)-\ln n\right]=\lim _{x \rightarrow+\infty} \frac{a_{1}^{x} \ln a_{1}+\cdots+a_{n}^{x} \ln a_{n}}{a_{1}^{x}+a_{2}^{x}+\cdots+a_{n}^{x}}=\ln \max \left\{a_{1}, a_{2} \cdots a_{n}\right\}$

所以原极限为$\max \left\{a_{1}, a_{2} \cdots a_{n}\right\}$

当$a=-\infty$时,由上述结论易知,原极限为$\min \left\{a_{1}, a_{2} \cdots a_{n}\right\}$

\section{找出错误：}
由 $\mathrm{L'Hospital}$ 法则知
$$
	\lim _{x \rightarrow \infty} \frac{x+\sin x}{x}=\lim _{x \rightarrow \infty} \frac{1+\cos x}{1}
$$

等式左侧
$$
	\lim _{x \rightarrow \infty} \frac{x+\sin x}{x}=\lim _{x \rightarrow \infty}\left(1+\frac{\sin x}{x}\right)=1
$$

等式右侧
$$
	\lim _{x \rightarrow \infty} \frac{1+\cos x}{1}=\lim _{x \rightarrow \infty}(1+\cos x)=1+\lim _{x \rightarrow \infty} \cos x
$$

因此
$$
	1=1+\lim _{x \rightarrow \infty} \cos x
$$

从而 $\lim \cos x=0$

洛必达需要极限存在.

\section{已知$f(x)=\left\{\begin{array}{ll}
		  \frac{g(x)}{x}, & x \neq 0 \\
		  0,              & x=0
	  \end{array}\right.$,其中$g(0)=0$, $g^{\prime}(0)=0$, $g^{\prime \prime}(0)=A$ .求$f^{\prime}(0)$}
\textbf{解}\quad
$$f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{g(x)}{x^{2}}=\lim _{x \rightarrow 0} \frac{g^{\prime}(x)}{2 x}=\lim _{x \rightarrow 0} \frac{g^{\prime \prime}(x)}{2}=\frac{A}{2}$$

\section{设函数 $f(x)$在$(a,+\infty)$可导,且$\lim _{x \rightarrow+\infty}\left[f^{\prime}(x)+k f(x)\right]=A$,证明$\lim _{x \rightarrow+\infty} f(x)=A$}
\textbf{证}\quad
$$
	\begin{aligned}
		\lim _{x \rightarrow+\infty} f(x) & =\lim _{x \rightarrow+\infty} \frac{\mathrm{e}^{k x} f(x)}{\mathrm{e}^{k x}}=\lim _{x \rightarrow+\infty} \frac{\mathrm{e}^{k x}\left[k f(x)+f^{\prime}(x)\right]}{k \mathrm{e}^{k x}} \\
		                                  & =\lim _{x \rightarrow+\infty} \frac{k f(x)+f^{\prime}(x)}{k}=\frac{1}{k} \lim _{x \rightarrow+\infty}\left[f^{\prime}(x)+k f(x)\right]=\frac{A}{k}
	\end{aligned}
$$

\section{设$f(x)$在点$x$处有二阶导数,求证: $f^{\prime \prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)+f(x-h)-2 f(x)}{h^{2}}$.由此推出结论:若$f(x)$ 是二阶可导的凸函数,必有$f^{\prime \prime}(x) \geqslant 0$.}
\textbf{解}\quad
$$
	\begin{array}{l}
		\lim _{h \rightarrow 0} \frac{f(x+h)+f(x-h)-2 f(x)}{h^{2}}                                                    \\
		=\lim _{h \rightarrow 0} \frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2 h}                                          \\
		=\lim _{h \rightarrow 0} \frac{f^{\prime}(x+h)-f^{\prime}(x)-\left[f^{\prime}(x-h)-f^{\prime}(x)\right]}{2 h} \\
		=\lim _{h \rightarrow 0} \frac{f^{\prime}(x+h)-f^{\prime}(x)}{2 h}+\lim _{h \rightarrow 0} \frac{f^{\prime}(x-h)-f^{\prime}(x)}{-2 h}=f^{\prime \prime}(x)
	\end{array}
$$

由此可知,$\forall x, \quad \frac{f^{\prime}(x+h)-f^{\prime}(x)}{h}>0$,所以由极限的保号性,$f^{\prime \prime}(x)=\lim _{h \rightarrow 0} \frac{f^{\prime}(x+h)-f^{\prime}(x)}{h}>0$

\section{由 Lagrange 中值定理我们知道 : 任取 $x \neq 0$,存在 $0<\theta_{x}<1$,使得$\mathrm{e}^{x}-1=x \mathrm{e}^{\theta_{x} x}$.证明 :$\lim _{x \rightarrow 0} \theta_{x}=\frac{1}{2}$.}
\textbf{证}\quad
$\lim _{x \rightarrow 0} \frac{e^{x}-1}{x e^{\theta_x x}}=1$

所以由洛必达定理$\lim _{x \rightarrow 0} \frac{e^{x}-1}{x e^{a x}}=\lim _{x \rightarrow 0}\frac{e^{x}}{e^{a x}(a x+1)}=\lim _{x \rightarrow 0}\frac{e^{(1-a) x}}{a x+1}=\lim _{x \rightarrow 0}\frac{(1-a) e^{0}}{a}=1$

解得$a=\frac{1}{2}$,即$\theta_x=\frac{1}{2}$.

\section{设$x_{1}=1, x_{n+1}=\ln \left(1+x_{n}\right), n=1,2, \cdots$求极限$\lim _{n \rightarrow \infty} n x_{n} $}
\textbf{解}\quad
由于$\ln(1+x)<x$,所以显然数列单减.数列显然由有下界$0$,所以由单调收敛原理可知,数列收敛.

设其极限为$A$,解得$A=\ln(1+A)$,所以$A=0$.

由$\mathrm{Stolz}$定理,$\lim _{x \rightarrow 0}nx_n=\lim _{n \rightarrow +\infty}\frac{1}{\frac{1}{x_n}-\frac{1}{x_{n-1}}}$

而$\lim _{x \rightarrow 0}\frac{-1}{\frac{1}{x}-\frac{1}{\ln(x+1)}}=2$

所以由海涅定理,原极限为2.

\end{document}